Question

From the top of a bridge that is 50m high, two boats can be seen anchored in a marina. One boat is anchored in the direction S20ºW, its angle of depression is 40º. The other boat is anchored in the same direction S60ºE, and its angle of depression is 30º. Determine the distance between the two boats.

Answer 1

Solution:

First we will draw the Diagram of the given Problem:

Representing the four directions as east, west, North And South

Let OB be the bridge = 50 m

OS represent one boat which is  anchored in the direction S 20º W.

Taking it's vertical component

So, OM= r sin 20°

Angle of Depression from top of the bridge to This ship = 40°

And OT represents another boat which is S 60º E.

Taking Vertical component of ON which is= P sin 60°

Angle of Depression from top of the bridge to This ship = 30°

In Δ B OM

$Tan 30= \frac{BO}{OM}\\\\ \frac{1}{\sqrt3}=\frac{50}{P Sin 60}\\\\ P = 100$ Use Sin 60 = $\frac{\sqrt3}{2}$

In Δ B ON

$Tan 40= \frac{BO}{ON}\\\\ 0.840=\frac{50}{r Sin 20}\\\\ r = 172.413$

As sin 20 = .342

Distance between two ships = d

Using law of cosines

$Cos 80 = \frac{P^2 + r^2-d^2}{2pr}\\\\ 0.18=\frac{(100)^2+(172.413)^2-d^2}{2 \times 100 \times 172.413}\\\\ 6206.76=39726.242 - d^2 \\\\d^2=33519.482 \\\\ d= 183.083$ meter